A common factor of `x ^(4) - 256, x ^(3) - 4x ^(2) + 3x - 12 and x ^(2) - 7x + 12 ` is

Options

A. `x + 4`
B. `x - 3`
C. `x -4`
D. `x + 3`

Answer

To find a common factor of the polynomials \( x^4 - 256 \), \( x^3 - 4x^2 + 3x - 12 \), and \( x^2 - 7x + 12 \), we will factor each polynomial and identify any common factors.

### Step 1: Factor \( x^4 - 256 \)

The expression \( x^4 - 256 \) can be recognized as a difference of squares:

\[ x^4 - 256 = (x^2)^2 - (16)^2 = (x^2 - 16)(x^2 + 16) \]

Next, we can further factor \( x^2 - 16 \) since it is also a difference of squares:

\[ x^2 - 16 = (x - 4)(x + 4) \]

Thus, the complete factorization of \( x^4 - 256 \) is:

\[ x^4 - 256 = (x - 4)(x + 4)(x^2 + 16) \]

### Step 2: Factor \( x^3 - 4x^2 + 3x - 12 \)

To factor \( x^3 - 4x^2 + 3x - 12 \), we can use the Rational Root Theorem or synthetic division. Testing \( x = 3 \):

\[ 3^3 - 4(3^2) + 3(3) - 12 = 27 - 36 + 9 - 12 = -12 \quad \text{(not a root)} \]

Testing \( x = 2 \):

\[ 2^3 - 4(2^2) + 3(2) - 12 = 8 - 16 + 6 - 12 = -14 \quad \text{(not a root)} \]

Testing \( x = 4 \):

\[ 4^3 - 4(4^2) + 3(4) - 12 = 64 - 64 + 12 - 12 = 0 \quad \text{(is a root)} \]

Now we can perform synthetic division of \( x^3 - 4x^2 + 3x - 12 \) by \( x - 4 \):

\[ \begin{array}{r|rrrr} 4 & 1 & -4 & 3 & -12 \\ & & 4 & 0 & 12 \\ \hline & 1 & 0 & 3 & 0 \\ \end{array} \]

This gives us:

\[ x^3 - 4x^2 + 3x - 12 = (x - 4)(x^2 + 3) \]

### Step 3: Factor \( x^2 - 7x + 12 \)

Next, we factor \( x^2 - 7x + 12 \):

\[ x^2 - 7x + 12 = (x - 3)(x - 4) \]

### Step 4: Identify Common Factors

Now we have the factorizations:

1. \( x^4 - 256 = (x - 4)(x + 4)(x^2 + 16) \)
2. \( x^3 - 4x^2 + 3x - 12 = (x - 4)(x^2 + 3) \)
3. \( x^2 - 7x + 12 = (x - 3)(x - 4) \)

The common factor in all three polynomials is \( x - 4 \).

### Final Answer

The common factor of \( x^4 - 256 \), \( x^3 - 4x^2 + 3x - 12 \), and \( x^2 - 7x + 12 \) is:

\[ \boxed{x - 4} \]